The equation of a circle $C$ is $x^2+y^2-18x+16y+141 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-18x) + (y^2+16y) = -141$ $(x^2-18x+81) + (y^2+16y+64) = -141 + 81 + 64$ $(x-9)^{2} + (y+8)^{2} = 4 = 2^2$ Thus, $(h, k) = (9, -8)$ and $r = 2$.